Calculating Acceleration and Work in a Changing Force System: A Graph Problem

[tachu101]

Homework Statement

This Question has a graph with it. (See Attachment)

An object of mass (m) moves in a straight line. The net force varies with the displacement. The mass starts at 0 meters and at 0 seconds.The mass goes a total distance of 20 meters. X1 is equal to 11 meters.

Mass --- .87kg
Net Force --- 4 Newtons
X1 --- 11 meters

Find the:
Acceleration of the object when its displacement is 11/3 meters?
The time taken for the object to be displaced from 0 to X1 (11 meters)?
Amount of work done by the net force in displacing the object from 0 to X1 (11 meters)?
The speed of the object at displacement X1 (11 meters)?
The final speed of the object at 20 meters?
The impulse experienced by the object as it is displaced from X1 to 20 meters (9 meters)?

Homework Equations

F=ma
Vf^2=Vo^2+2as
W=Fd

The Attempt at a Solution

I am not sure if I am doing this right, can someone make sure?

For the acceleration would it be ---- F=ma so --- 4N=.87kg(a) ---- a= 4.597m/sec

For the speed of the object at displacement X1 (11 meters) I used
Vf^2=Vo^2+2as ----- Vf^2=(0)+2(4.597m/s)(11meters) ---- Vf= 10.1 m/s

For the time I used --- t=d/(average velocity) ---- Average velocity= (initial+final/2) =
(0+10.1/2) = 5.05 m/s ? (I am not sure about this)... Then I put it into t=d/(average velocity) ----- t=(11 meters)/(5.05m/s) = 2.19 seconds? (check please)

For the work W=Fd ---- W=(4)(11) ---- 44J

I am not sure how to get the overall final speed of the mass because the force is changing at the end.

I have no idea how to do the impulse experienced by the object from X1 (11 meters) to 20 meters, I am not sure what the equation is.

Can you tell me if I am doing this correctly and how I should try to get the last two parts.

 

[azatkgz]

Cannot see the picture.It says "Atachment pending approval".

 

[tachu101]

I have the graph up now.

 

[azatkgz]

For time
x=v_0t+\frac{1}{2}at^2

t=\sqrt{\frac{2x}{a}

To find overall velocity:
Calculate the area under the graph.Area=\frac{1}{2}mv_f^2

Impulse:

I=m(v_f-v_{x_1})

v_f is the velocity at x=20 m

v_{x_1} is the velocity at x=11 m

 

[azatkgz]

Other solutions seems to be right.

 

[tachu101]

I do not understand how the area under the graph could get me the final velocity, can anyone explain?

 

[azatkgz]

Because area under the F-x graph equals to work.Here work goes only for kinetic enrgy.

 

[tachu101]

So is the final velocity 11.9 m/s

 

[azatkgz]

Yes,you're right!